We need to solve the equation $|x-2| + |x-3| + 2|x-4| = 9$.

1. Problem Description

We need to solve the equation $|x-2| + |x-3| + 2|x-4| = 9$.

2. Solution Steps

The critical points are $x=2$, $x=3$, and $x=4$. We will consider the following cases:

Case 1: $x \leq 2$

In this case, $x-2 \leq 0$, $x-3 \leq 0$, and $x-4 \leq 0$. Thus $|x-2| = 2-x$, $|x-3| = 3-x$, and $|x-4| = 4-x$.

The equation becomes:

$(2-x) + (3-x) + 2(4-x) = 9$

$2-x+3-x+8-2x = 9$

$13-4x = 9$

$4x = 4$

$x=1$

Since $1 \leq 2$, $x=1$ is a valid solution.

Case 2: $2 < x \leq 3$

In this case, $x-2 > 0$, $x-3 \leq 0$, and $x-4 \leq 0$. Thus $|x-2| = x-2$, $|x-3| = 3-x$, and $|x-4| = 4-x$.

The equation becomes:

$(x-2) + (3-x) + 2(4-x) = 9$

$x-2+3-x+8-2x = 9$

$9-2x = 9$

$2x = 0$

$x=0$

Since $2 < x \leq 3$, $x=0$ is not a valid solution.

Case 3: $3 < x \leq 4$

In this case, $x-2 > 0$, $x-3 > 0$, and $x-4 \leq 0$. Thus $|x-2| = x-2$, $|x-3| = x-3$, and $|x-4| = 4-x$.

The equation becomes:

$(x-2) + (x-3) + 2(4-x) = 9$

$x-2+x-3+8-2x = 9$

$3 = 9$

This is a contradiction, so there are no solutions in this case.

Case 4: $x > 4$

In this case, $x-2 > 0$, $x-3 > 0$, and $x-4 > 0$. Thus $|x-2| = x-2$, $|x-3| = x-3$, and $|x-4| = x-4$.

The equation becomes:

$(x-2) + (x-3) + 2(x-4) = 9$

$x-2+x-3+2x-8 = 9$

$4x - 13 = 9$

$4x = 22$

$x = \frac{22}{4} = \frac{11}{2} = 5.5$

Since $5.5 > 4$, $x=5.5$ is a valid solution.

3. Final Answer

The solutions are $x=1$ and $x=5.5$.